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aleksley [76]
3 years ago
13

Two pressure gauges measure a pressure drop of 16.3 psi (lb/in.2) at the entrance and exit of an old buried pipeline. The origin

al drawings have been lost. If the 6-in. galvanized iron pipe carries water at 68°F with a flow rate of 1.64 cfs (ft3/s), determine the length of the horizontal underground line ignoring minor losses.
Engineering
1 answer:
Sholpan [36]3 years ago
7 0

Answer: 866.25ft

Explanation:

Pressure drop= 16.3 psi

1 feet of water column= 0.4335psi

x= 16.3psi

Therefore, head loss of water= 16.3/0.4335=37.6ft

In inches, head loss of water= 37.6 × 12=451.2 inches

Q= 1.64ft^2/s

A= pi/4 D^2

Where D= 6 inches, 0.5ft

A= 3.142/4 × (0.5)^2

A= 0.1962ft^2

V= Q/A= 1.64/ 0.1962

V= 8.35ft/s, 100.28 in/s

REGNOLD'S NO

Re= SvD/ μ=

Where, D= 6 inches, V= 100.28 inches/s, S=62.4lb/ft^3= 0.03611lb/inches^3

At 68F, Dynamic viscosity μ= 1.0016× 10^-3NS/m^2

1psi= 6895N/m^3

μ= 1.0016× 10^-3NS/m^2/6895N/m^3

μ= 1.452× 10^-7lbs/ins^2

Re= 0.03611× 100. 28×6/1.42×10^-7

Re= 1.49×10^8

e/D----- Relative roughness

e= 0.0005ft

e/D= 0.001

Therefore, for Re= 1.49×10^8 and e/D= 0.001

F=0.02

Head loss is given by Dancy-Weisbach formula

hL= fLV^2/2gD

Wherr g= 9.81m/s^2= 386.12inches/s^2

451.2= 0.02×L×(100.28^2)/2 × 386.12inches/s^2× 6

L= 10395inches

L= 10395/12= 866.25ft

Therefore, the length of pipe is 866.25ft.

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A water skier leaves the end of an 8 foot tall ski ramp with a speed of 20 mi/hr and at an angle of 250. He lets go of the tow r
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At highest point:

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When he lands:

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v0 = (26.5*i + 12.4*j) ft/s

I set up the coordinate system with the origin at the base of the ramp under its end, so:

R0 = (0*i + 8*j) ft

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X(t) = X0 + Vx0 * t

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Vx(t) = Vx0

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In the heighest point of the trajectory the vertical speed will be zero because that is the point where it transitions form going upwards (positive vertical speed) to going down (negative vertical speed), and it crosses zero.

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